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\title{第二周理论作业}

\author{徐晟 \\ 信計 3200104397}

\begin{document}

\section{1.8.1.1}
the width of interval at the nth step is
$$2*(\frac{1}{2})^{n}=(\frac{1}{2})^{n-1}$$
and the maximum possible distance is 1.
\section{1.8.1.2}
因为 $$ \frac{b_{0}-a_{0}}{a_{0}}*(\frac{1}{2})^{n} \leq \epsilon $$
所以 $$ n \geq \frac{\lg(b_{0}-a_{0})-\lg \epsilon-\lg a_{0}}{\ln 2}-1 $$
\section{1.8.1.3}

\begin{tabular}{cc}
\hline
$x_{0}$&$x_{1}$&$x_{2}$&$x_{3}$&$x_{4}$\\
\hline
-1.000&-0.8125&-0.7708&-0.7688&-0.7688\\
\end{tabular}

\section{1.8.1.4}
令 $e_{n}= |x_{n}-\alpha|$
$$ \frac{e_{n+1}}{e_{n}^{2}} = \frac{f^{''}(\xi)}{2*f^{'}(x_{n})}$$
其中$\xi$ is between $\alpha$ and $x_{n}$
所以 $ s=2, c=\frac{f^{''}(\xi)}{2*f^{'}(x_{n})} $
\section{1.8.1.5}
因为 
$$ f^{'}(x)=\frac{1}{1+x^{2}} \textgreater 0 $$
由Corollary 1.41得其不收敛
\section{1.8.1.6}
$$ \frac{x_{n+1}-x_{n}}{x_{n}-x_{n-1}} = -x_{n+1}*x_{n} $$
$$ |\frac{x_{n+1}-x_{n}}{x_{n}-x_{n-1}}| \leq \frac{1}{p^{2}} \textless 1 $$
the sequence of values converges.
\section{1.8.1.7}
由于 0 is between $a_{0}$ and $b_{0}$
$$ n \geq \frac{\lg(b_{0}-a_{0})-\lg \epsilon-\lg (max(|a_{0}|,|b_{0}|)) }{\ln 2}-1 $$
所以原误差估计不一定成立

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